The unit circle is the map that connects angles to coordinates, and coordinates to the trigonometric functions. Once you understand it, you can find the sine, cosine, and tangent of any angle — positive, negative, or larger than $360^\circ$ — without a calculator. In this lesson, you will learn to navigate this map like a pro.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Imagine a circle with radius 1 centred at the origin. If you walk around the circumference to an angle of $45^\circ$ ($\frac{\pi}{4}$ radians), what are your $x$- and $y$-coordinates? How do these coordinates relate to $\sin 45^\circ$ and $\cos 45^\circ$?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: tan(θ) = sin(θ) + cos(θ).
Right: tan(θ) = sin(θ)/cos(θ); it is a ratio, not a sum.
📚 Core Content
The unit circle is a circle with radius $1$ centred at the origin $(0, 0)$ of the coordinate plane. It is the foundation of trigonometry because it connects angles to coordinates in a simple, universal way.
If you start at the point $(1, 0)$ and rotate anticlockwise by an angle $\theta$, the point where you land on the unit circle is:
$$P(\theta) = (\cos \theta, \sin \theta)$$
This means:
Since the unit circle has radius $1$ and centre $(0, 0)$, its equation is:
$$x^2 + y^2 = 1$$
Substituting $x = \cos \theta$ and $y = \sin \theta$ gives the fundamental identity:
$$\cos^2 \theta + \sin^2 \theta = 1$$
Not every angle lands in Quadrant I. The ASTC rule tells you which trigonometric ratios are positive in each quadrant:
| Quadrant | Positive ratios | Memory aid |
|---|---|---|
| I ($0$ to $\frac{\pi}{2}$) | All ($\sin, \cos, \tan$) | All |
| II ($\frac{\pi}{2}$ to $\pi$) | Sin only | Sine |
| III ($\pi$ to $\frac{3\pi}{2}$) | Tan only | Tangent |
| IV ($\frac{3\pi}{2}$ to $2\pi$) | Cos only | Cosine |
The reference angle $\alpha$ is the acute angle that the terminal side makes with the $x$-axis. It is always positive and between $0$ and $\frac{\pi}{2}$.
| Quadrant | Reference angle formula |
|---|---|
| I | $\alpha = \theta$ |
| II | $\alpha = \pi - \theta$ |
| III | $\alpha = \theta - \pi$ |
| IV | $\alpha = 2\pi - \theta$ |
Once you know the reference angle, you can find the exact trig value using special triangles, then apply the correct sign from ASTC.
🧮 Worked Examples
🧪 Activities
1 $\theta = \pi$
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2 $\theta = \frac{5\pi}{4}$
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3 $\theta = -\frac{\pi}{2}$
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4 $\theta = \frac{5\pi}{6}$
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1 $\sin \frac{4\pi}{3}$
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2 $\cos \frac{7\pi}{6}$
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3 $\tan \frac{11\pi}{6}$
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4 $\cos\left(-\frac{2\pi}{3}\right)$
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Earlier you were asked: What are your coordinates at $45^\circ$ on the unit circle, and how do they relate to $\sin 45^\circ$ and $\cos 45^\circ$?
At $\theta = \frac{\pi}{4}$ ($45^\circ$), the point on the unit circle is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. By the unit circle definition, the $x$-coordinate is $\cos \theta$ and the $y$-coordinate is $\sin \theta$. Therefore:
This is why the exact values for $45^\circ$ are so memorable: on the unit circle, the $x$ and $y$ coordinates are equal because the angle bisects the first quadrant.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. (a) State the exact coordinates of the point on the unit circle corresponding to $\theta = \frac{3\pi}{4}$. (b) Hence, write down the exact values of $\sin \frac{3\pi}{4}$ and $\cos \frac{3\pi}{4}$. 3 MARKS
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Answer in your workbook.
9. If $\cos \theta = -\frac{5}{13}$ and $\theta$ is in Quadrant II, find the exact value of $\sin \theta$. Show your working. 2 MARKS
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Answer in your workbook.
10. A student claims that $\sin \theta$ is always positive when $\theta$ is between $0$ and $\pi$, and always negative when $\theta$ is between $\pi$ and $2\pi$. Evaluate this claim, using specific examples from the unit circle to support your answer. 3 MARKS
Type your answer below:
Answer in your workbook.
1. $(-1, 0)$
2. $\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$
3. $(0, -1)$
4. $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
1. $\sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$ (QIII, reference $\frac{\pi}{3}$)
2. $\cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$ (QIII, reference $\frac{\pi}{6}$)
3. $\tan \frac{11\pi}{6} = -\frac{\sqrt{3}}{3}$ (QIV, reference $\frac{\pi}{6}$)
4. $\cos\left(-\frac{2\pi}{3}\right) = -\frac{1}{2}$ (QIII, reference $\frac{\pi}{3}$)
1. A — $\frac{\pi}{2}$ corresponds to $(0, 1)$.
2. A — $\cos \theta = x$-coordinate on the unit circle.
3. A — Point is $(0, -1)$, so $\sin = -1$.
4. A — Both $x$ and $y$ are negative in QIII.
5. A — $\sin^2 \theta = 1 - \frac{16}{25} = \frac{9}{25}$; QII $\to$ positive $\to \frac{3}{5}$.
Q8 (3 marks): (a) $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ [1]. (b) $\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$ [1], $\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$ [1].
Q9 (2 marks): $\sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}$ [1]. In QII, sine is positive, so $\sin \theta = \frac{12}{13}$ [1].
Q10 (3 marks): The claim is partially correct but imprecise [0.5]. Between $0$ and $\pi$, sine is positive in QI and QII, so the first part is correct [0.5]. However, between $\pi$ and $2\pi$, sine is negative in QIII and QIV, so the second part is also correct [0.5]. But the claim ignores that $\sin \pi = 0$ and $\sin 2\pi = 0$, which are neither positive nor negative [1]. Better answer: the student's claim is broadly true for the open intervals $(0, \pi)$ and $(\pi, 2\pi)$, but false at the endpoints where sine is zero [0.5].
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